Abstract

Ran and Reurings (2004) established an interesting analogue of Banach Contraction Principle in a complete metric space equipped with a partial ordering and also utilized the same oneto discuss the existence of solutions to matrix equations. Motivated by this paper, we prove results on coincidence points for a pair of weakly increasing mappings satisfying a nonlinear contraction condition described by a rational expression on an ordered complete metric space. The uniqueness of common fixed point is also discussed. Some examples are furnished to demonstrate the validity of the hypotheses of our results. As an application, we derive an existence theorem for the solution of an integral equation.

1. Introduction with Preliminaries

A variety of generalizations of the Classical Banach Contraction Principle [1] are available in the existing literature of metric fixed point theory. The majority of these generalizations are obtained by improving the underlying contraction condition (e.g., [2]). Presently, there is vigorous research activity to prove existence results on complete metric spaces equipped with a partial ordering. In fact, various existence and uniqueness theorems on fixed and common fixed point for monotone mappings are of paramount importance in the study of nonlinear equations which generate natural interest to establish usable fixed point theorems in partial metric spaces (e.g., [1–24]).

Very recently, Harjani et al. [25] proved a fixed point theorem in partially ordered metric spaces satisfying a contractive condition of rational type due to Jaggi [26]. The aim of this paper is to prove some results of Harjani et al. [25] type for a pair of self-mappings. We accomplish this using the concept of weakly increasing property due to Nashine and Samet [14] (also see [4, 27, 28]). Some examples are also furnished to demonstrate the validity of the hypotheses of our results. As an application, we establish the existence of solution to an integral equation (also see [2, 23, 29, 30]).

Before presenting our results, we recall some notations, definitions, and examples required in our subsequent discussions.

Definition 1. Let be a nonempty set. Then is called an ordered (partial) metric space if
(i) is a partially ordered set and (ii) is a metric space.

Definition 2. Let be a partially ordered set. Then(a)elements are called comparable with respect to “” if either or ; (b)a mapping is called nondecreasing with respect to “” if implies .

Let be a nonempty set and be a given mapping. For every , we denote by , the subset of defined by

Definition 3. Let be a partially ordered set and let be two mappings such that . We say that is weakly increasing with respect to if and only if for all , one has:

Remark 4. If is the identity mapping on , then is weakly increasing with respect to if and only if for all .

As mentioned earlier, the notion of weakly increasing mappings was introduced in [4] which is presently in use (e.g. [27, 28]). In what follows, we furnish a relatively new example to demonstrate the preceding definition.

Example 5. Consider endowed with the natural order . Define two mappings as
In order to show that the mapping is weakly increasing with respect to mapping , we distinguish three cases.Firstly, we consider the case . Let , that is, so that and henceforth or . By the definitions of , we have
Secondly, we argue the case . Let , that is, , which amounts to say (in view of definition ) that or , implying thereby
Finally, we need to consider the case . Let , that is, . By the definition of , we have , so that . Now, in view of definition of , we have , so that , yielding thereby . Now, we have
Thus we have shown that is weakly increasing with respect to .

Definition 6. Let be an ordered metric space. We say that is regular if and only if the following hypothesis holds:
if is a nondecreasing sequence in with respect to such that , then for all .

Definition 7. A pair of self-mappings of a metric space is said to be compatible if and only if whenever is a sequence in such that

Definition 8 (see [18]). A pair of self-mappings of a metric space is said to be reciprocally continuous if and only if and for every sequence in satisfying
for some .Notice that a pair of continuous mappings is always reciprocally continuous but not conversely as substantiated by examples in [18].

Definition 9 (see [19]). A pair of self mappings of a metric space is said to be weakly reciprocally continuous if and only if , for every sequence in satisfying
for some .Evidently, every pair of reciprocally continuous mappings is always weakly reciprocally but not conversely as demonstrated in Pant et al. [19].

2. Results

The main result of this paper runs as follows

Theorem 10. Let be a partially ordered set equipped with a metric on such that is a complete metric space. Let be two mappings satisfying (for pairs wherein and are comparable)
where , are nonnegative real numbers with . Suppose that (a) is regular and is weakly increasing with , (b)the pair is commuting as well as weakly reciprocally continuous. Then and have a coincidence point. That is, there exists such that .

Proof. Let be an arbitrary point in . Since , one can inductively construct a sequence in defined by
As and , using weakly increasing property of with respect to , we obtain
Continuing this process indefinitely, we get
Now, we proceed to show that for ,
As , using (9), we have
which implies that
so that
Let (13) holds for all . As , using (9), we have
so that
Now with , for any , we have
As , we have , that is, is Cauchy sequences in the complete metric space and hence there exists some such that
Owing to commutativity of with and by using (10) (for each ), we have
Since is weakly increasing with , we can write
so that is nondecreasing. As the maps and are weakly reciprocally continuous, , which together with regularity of gives rise to ; that is, and are comparable. On using condition (9), we have
On making in the preceeding inequality, one gets
so that . Thus, we have shown that and have a coincidence point. This completes the proof.

Corollary 11. Let be a partially ordered set on which there is a metric on such that is a complete metric space. Let be given mappings satisfying (for pairs wherein and are comparable)
where is a nonnegative real number with . Suppose that (a) is regular and is weakly increasing with , (b)the pair ( is commuting as well as weakly reciprocally continuous. Then and have a coincidence point. That is, there exists such that .

Theorem 12. Let be a partially ordered set equipped with a metric on such that is a complete metric space. Let be two mappings satisfying (for pairs wherein and are comparable)
where , are non-negative real numbers with . Suppose that (a) is weakly increasing with , (b)the pair is compatible and reciprocally continuous. Then and have a coincidence point. That is, there exists such that .

Proof. Proceeding on the lines of the proof of Theorem 10, one can furnish a sequence such that
Now, it remains to show that is a coincidence point of and . To accomplish this, as the pair is compatible as well as reciprocally continuous,
whenever
By using (29) in (28), we have , so that .

By appealing Theorems 10 and 12, one can also have the following natural theorem.

Theorem 13. Let be a partially ordered set on which there is a metric on such that is a complete metric space. Let be nondecreasing mapping satisfying (for pairs wherein and are comparable)
where , are non-negative real numbers such that . Suppose that (I) for all ,(II)either is continuous or is regular. Then, has a fixed point.

Proof. The proof of this theorem can be outlined on the lines of the proof of Theorem 10 realizing to be the identity mapping on .

Now, we introduce the following property which will be utilized in our next theorem.

Property (A). If is a nondecreasing sequence in such that , then is comparable to , for all .

Theorem 14. Let be a partially ordered set on which there exists a metric on such that is a complete metric space. Let be given mappings satisfying (for pairs wherein and are comparable)
where , are non-negative real numbers with . Suppose that (a) is regular and is weakly increasing with , (b)the pair ( is commuting as well as weakly reciprocally continuous, (c) satisfies Property (A). Then and have a common fixed point.

Proof. Proceeding on the lines of the proof of Theorem 10, one can inductively construct nondecreasing sequence , such that , and . Since and are comparable (for all ), by using condition (32) we have
Taking the limit as , one gets
This completes the proof.

Theorem 15. Let be a partially ordered set on which there exists a metric on such that is a complete metric space. Let be given mappings satisfying (for pairs wherein and are comparable)
where , are non-negative real numbers with . Suppose that (a) is weakly increasing with , (b)the pair is compatible and reciprocally continuous, (c) satisfies Property (A).
Then and have a common fixed point.

3. Uniqueness Results

In what follows, we investigate the conditions under which Theorem 10 ensures the uniqueness of common fixed point.

Theorem 16. If, in addition to the hypotheses of Theorem 10, every , there exists a such that is upper bound of and , then and have a unique common fixed point.

Proof. In view of Theorem 10, the set of coincidence points of the maps and is non-empty. If and are two coincidence points of the maps and (i.e., , ), then we proceed to show that
In view of the additional hypothesis of this theorem, there exists such that is upper bound of and . Put , and choose such that . Now, proceeding on the lines of the proof of Theorem 10, one can inductively define sequence such that
wherein
Further, setting , and , one can also define the sequences and such that
For every , we have
Since is upper bound of and , then
It is easy to show that and for all , then and are comparable with . On using (9), we have
or
Owing to (43), we can write
Taking the limit as in (44), we get
as .Similarly, one can also show that
On using (45) and (46), we can have
so that , that is, . Thus, we have proved (36).Since , owing to commutativity of and , one can write
which on inserting gives rise to
Thus, is another coincidence point of the pair. Now, due to (for every coincidence point and ) and owing to the fact that is coincidence point of the pair , it follows that , that is,
On making use of (49) and (50), we can have
which shows that is a common fixed point of and .To prove the uniqueness, let be another common fixed point of the pair . Since and are coincidence points, then .Owing to that and are common fixed point of the pairs , one can have
This completes the proof.

Example 17. Consider equipped with natural order and . Thus , is a partially ordered set wherein the two elements are not comparable to each other. Also, is a complete metric space under Euclidean metric. Defined mappings and as
As and in are merely comparable to themselves, inequality (9) is vacuously satisfied for every .Notice that . Also, for ,
so that
Otherwise, for ,
so that
Thus is weakly increasing with respect to .If is a nondecreasing sequence converging to , then necessarily must be constant sequence, that is, , for all , so that limit is an upper bond for all the terms in the sequence which shows that is regular.By using definitions of the maps and , we have
wherein so that
which shows that the pair is weakly reciprocally continuous. Also, the pair is clearly commuting.Now, we show that for every and in , there exists a such that is upper bound of and . If , then choice of is obvious. Otherwise, if and , we can choose such that is upper bound of and .Thus, we have shown that all the conditions of Theorem 16 are satisfied and is the unique common fixed point of and .

Theorem 18. If, in addition to the hypotheses of Theorem 12, for pairs , there exists a such that is upper bound of and , then and have a unique common fixed point.

Corollary 19. In addition to the hypotheses of Theorem 16 (or Theorem 18), suppose that for every there exists a such that is upper bound of and . Then has a unique fixed point; that is, there exists a unique such that .

Proof. In Theorem 16, if (the identity mapping on ), we have the result.

Example 20. Consider equipped with the usual metric and natural order . Define two mappings by
Then, evidently is a partially ordered set and the maps and satisfy inequality (9) with and , for .To show that is weakly increasing with respect to , notice that . Firstly, we argue the case . Let , that is, so that or . Using definitions of and , we have
Secondly, if , then we have . By using definition of , we have , so that
Finally, we consider the case . Let , that is, . In view of definition of , we have , so that . By using the definition of , we have , so that or . Thus, we have
Therefore is weakly increasing with respect to .Since and , are continuous, therefore this pair of maps is reciprocally continuous.Now, we show that maps and , are compatible. If , and , then and henceforth
implying thereby so that and , are compatible.Thus, we have shown that all the conditions of Theorem 18 are satisfied and is the unique fixed point of and .

4. An Application

In this section, we present an application of Theorem 13 and used the idea of Ćirić et al. [29] to define a partial order and prove an existence theorem for the solution of an integral equation.

Theorem 21. Consider the integral equation
with wherein(i)the functions and are continuous,(ii)for each ,
where denotes a partial order relation on , (iii)there exists a continuous function such that
for each and is also comparable ,(iv). Then the integral equation (66) has a unique solution in .

Proof. Consider with the usual supremum norm, that is,
for . Define on a partial order as follows :
Then is a partially ordered set and is a complete metric space.Moreover for any increasing sequence in converging to , we have for any . Also for every there exists which depends on and and is also comparable to and (cf. [21]).Define by
Let us prove that
Let . From (ii), for all , we have
Then (72) holds.Now, for all with , we have (by (iii))
Hence
On the other hand, it is demonstrated in [16] that condition of regularity is satisfied for .Thus, all the hypotheses of Theorem 13 are satisfied for , and then has a fixed point ; that is, is a solution of the integral equation (66).

Acknowledgments

The authors would like to thank the referees for careful reading and for providing valuable suggestions and comments for this paper. Also, Poom Kumam would like to thank the King Mongkuts University of Technology Thonburi (KMUTT) for the financial support and Dhananjay Gopal is thankful to CSIR, Government of India, Grant no. 25(0215)/13/EMR-II.